\newproblem{lay:6_2_25}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.2.25}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Prove the following theorem:
	
	Let $U\in\mathcal{M}_{n\times n}$ be an orthonormal matrix and $\forall\mathbf{x},\mathbf{y}\in\mathbb{R}^n$, then
	\begin{enumerate}[a.]
		\item $\|U\mathbf{x}\|=\|\mathbf{x}\|$
		\item $(U\mathbf{x})\cdot(U\mathbf{y})=\mathbf{x}\cdot\mathbf{y}$
		\item $(U\mathbf{x})\cdot(U\mathbf{y})=0 \Leftrightarrow \mathbf{x}\cdot\mathbf{y}=0$
	\end{enumerate}
}{
   % Solution
	Let's prove first point b:
	\begin{center}
		$\begin{array}{rcl}
			(U\mathbf{x})\cdot(U\mathbf{y})&=&(U\mathbf{x})^T(U\mathbf{y})=\mathbf{x}^TU^TU\mathbf{y}=\mathbf{x}^\mathbf{y}=\mathbf{x}\cdot\mathbf{y}
		\end{array}$
	\end{center}
	where we have made used that for any orthonormal matrix $U^TU=I$.
	
	Let's prove now point a:
	\begin{center}
		$\begin{array}{rcl}
			\|U\mathbf{x}\|^2&=&(U\mathbf{x})\cdot(U\mathbf{x})=\mathbf{x}\cdot\mathbf{x}=\|\mathbf{x}\|^2
		\end{array}$
	\end{center}
	Taking the square root
	\begin{center}
		$\begin{array}{rcl}
			\|U\mathbf{x}\|&=&\|\mathbf{x}\|
		\end{array}$
	\end{center}
	
	Finally, point c. From point b we know that $(U\mathbf{x})\cdot(U\mathbf{y})=\mathbf{x}\cdot\mathbf{y}$. So, it is obvious that
	$(U\mathbf{x})\cdot(U\mathbf{y})=0$ iff $\mathbf{x}\cdot\mathbf{y}=0$.
}
\useproblem{lay:6_2_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
